Lời giải:
Đặt $2^x=a$.
PT $\Leftrightarrow (2^x)^2-10.2^x+16=0$
$\Leftrightarrow a^2-10a+16=0$
$\Leftrightarrow a^2-2a-8a+16=0$
$\Leftrightarrow a(a-2)-8(a-2)=0$
$\Leftrightarrow (a-8)(a-2)=0$
$\Rightarrow a=8$ hoặc $a=2$
Nếu $a=2\Leftrightarrow 2^x=2=2^1\Rightarrow x=1$
Nếu $a=8\Leftrightarrow 2^x=8=2^3\Rightarrow x=3$
Ta có : \(4^x-10.2^x+16=0\)
=> \(\left(2^x\right)^2-2^x.2.5+25-9=0\)
=> \(\left(2^x-5\right)^2-3^2=0\)
=> \(\left(2^x-5-3\right)\left(2^x-5+3\right)=0\)
=> \(\left(2^x-8\right)\left(2^x-2\right)=0\)
=> \(\left[{}\begin{matrix}2^x-8=0\\2^x-2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2^x=8\\2^x=2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = 3, x = 1 .
